markilue
/
self_example
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

leecode更新

This commit is contained in:
markilue 2022-10-14 17:32:03 +08:00
parent 8d8994786c
commit 5de2a0889b
3 changed files with 506 additions and 78 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

195
Leecode/src/main/java/com/markilue/leecode/backtrace/combinationSum2.java Normal file
View File

@ -0,0 +1,195 @@
package com.markilue.leecode.backtrace;
import org.junit.Test;
import java.util.*;
/**
* @BelongsProject: Leecode
* @BelongsPackage: com.markilue.leecode.backtrace
* @Author: markilue
* @CreateTime: 2022-10-14 09:49
* @Description: TODO 力扣40题 组合总和II：
* 给定一个候选人编号的集合 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
* candidates 中的每个数字在每个组合中只能使用 一次 。
* 注意：解集不能包含重复的组合。
* @Version: 1.0
*/
public class combinationSum2 {
@Test
public void test(){
int[] candidates = {10,1,2,7,6,1,5};
int target = 8;
System.out.println(combinationSum22(candidates, target));
}
@Test
public void test1(){
int[] candidates = {2,5,2,1,2};
int target = 5;
System.out.println(combinationSum2(candidates, target));
}
@Test
public void test2(){
int[] candidates = {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
int target = 30;
System.out.println(combinationSum21(candidates, target));
}
/**
* 回溯算法：这题的核心难点在于解集不能包含重复的组合，而候选者集合中会出现重复数字
* 1）这里考虑先将候选者集合排序,在进行逻辑处理
* 2）使用hashset去重
*
* @param candidates
* @param target
* @return
*/
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
backtracking(candidates, target, 0);
ArrayList<List<Integer>> lists = new ArrayList<>();
lists.addAll(result);
return lists;
}
List<List<Integer>> result = new ArrayList<>();
HashSet<List<Integer>> result1 = new HashSet<>();
List<Integer> cur = new ArrayList<>();
int sum = 0;
//当重复元素较多时，直接超出时间限制
public void backtracking(int[] candidates, int target, int val) {
if (sum == target) {
cur.sort(new Comparator<Integer>() {
@Override
public int compare(Integer o1, Integer o2) {
return o1-o2;
}
});
result1.add(new ArrayList<>(cur));
return;
}
for (int i = val; i < candidates.length && sum + candidates[i] <= target; i++) {
sum += candidates[i];
cur.add(candidates[i]);
backtracking(candidates, target, i + 1);
cur.remove(cur.size() - 1);
sum -= candidates[i];
}
}
/**
* 速度击败99.77%，内存击败52.35%
* @param candidates
* @param target
* @return
*/
public List<List<Integer>> combinationSum21(int[] candidates, int target) {
Arrays.sort(candidates);
boolean[] used = new boolean[candidates.length];
// System.out.println(Arrays.toString(used));
backtracking1(candidates, target, 0,used);
return result;
}
//代码随想录方法回溯：在回溯过程中就可以回溯，元素在同一个组合内是可以重复的，
// 但两个组合不能重复，所以需要去重同一树层上使用过的元素，同一树枝上不用去重
public void backtracking1(int[] candidates, int target, int val,boolean[] used) {
if (sum == target) {
result.add(new ArrayList<>(cur));
return;
}
for (int i = val; i < candidates.length && sum + candidates[i] <= target; i++) {
//如果used[i-1]==True，则说明同一树枝使用过candidates[i]
//如果used[i-1]=false,则说明同一树层使用过candidates[i]
//TODO 为什么同一树层使用过，就一定可以保证可以不在使用这个数？
// 因为由于这个树层使用第2数时，实际上等同于这个树层使用第1数，下个树层使用第2数的子集，所以一定是包含关系，可以直接跳过
if(i>0&&candidates[i]==candidates[i-1]&&used[i-1]==false){
continue;
}
//使用以下判断，也可以进行跳过
// if(i>val&&candidates[i]==candidates[i-1])
sum += candidates[i];
cur.add(candidates[i]);
used[i]=true;
backtracking1(candidates, target, i + 1,used);
used[i]=false;
cur.remove(cur.size() - 1);
sum -= candidates[i];
}
}
List<int[]> freq = new ArrayList<int[]>();
List<List<Integer>> ans = new ArrayList<List<Integer>>();
List<Integer> sequence = new ArrayList<Integer>();
/**
* 官方回溯方法：
* 实际上是将问题转为同一个数能无限使用(最多使用rest/num=>most次)的组合
* 与本人的combinationSum中的自己实现思路类似
* @param candidates
* @param target
* @return
*/
public List<List<Integer>> combinationSum22(int[] candidates, int target) {
Arrays.sort(candidates);
//这里是使用freq记录每个数字出现的次数
for (int num : candidates) {
int size = freq.size();
if (freq.isEmpty() || num != freq.get(size - 1)[0]) {
freq.add(new int[]{num, 1});
} else {
++freq.get(size - 1)[1];
}
}
//开始递归回溯
dfs(0, target);
return ans;
}
public void dfs(int pos, int rest) {
//等于0，意味着找到了这个序列
if (rest == 0) {
ans.add(new ArrayList<Integer>(sequence));
return;
}
//遍历到了最后，或者当前freq大于rest就返回
if (pos == freq.size() || rest < freq.get(pos)[0]) {
return;
}
//还不够就继续找
dfs(pos + 1, rest);
//同一个数最多能使用多少次
int most = Math.min(rest / freq.get(pos)[0], freq.get(pos)[1]);
for (int i = 1; i <= most; ++i) {
sequence.add(freq.get(pos)[0]);
dfs(pos + 1, rest - i * freq.get(pos)[0]);
}
for (int i = 1; i <= most; ++i) {
sequence.remove(sequence.size() - 1);
}
}
}

203
Leecode/src/main/java/com/markilue/leecode/backtrace/partition.java Normal file
View File

@ -0,0 +1,203 @@
package com.markilue.leecode.backtrace;
import org.junit.Test;
import java.util.ArrayList;
import java.util.List;
/**
* @BelongsProject: Leecode
* @BelongsPackage: com.markilue.leecode.backtrace
* @Author: markilue
* @CreateTime: 2022-10-14 11:40
* @Description: TODO 力扣131题 分割回文串：
* 给你一个字符串 s，请你将 s 分割成一些子串，使每个子串都是 回文串 。返回 s 所有可能的分割方案。
* 回文串 是正着读和反着读都一样的字符串。
* @Version: 1.0
*/
public class partition {
@Test
public void test() {
String s = "aab";
System.out.println(partition1(s));
}
@Test
public void test1() {
String s = "abba";
System.out.println(partition1(s));
}
@Test
public void test2() {
String s = "cdd";
System.out.println(partition2(s));
}
List<List<String>> result = new ArrayList<>();
List<String> cur = new ArrayList<>();
/**
* 自己的思路：分别按不同长度对s进行划分，不是回溯法
* 存在问题：同同一次划分时的长度可能不一样
*
* @param s
* @return
*/
public List<List<String>> partition(String s) {
//按i对s进行划分
for (int i = 1; i <= s.length(); i++) {
StringBuilder builder = new StringBuilder();
for (int j = 0; j < s.length(); j++) {
if (j != 0 && j % i == 0) {
cur.add(builder.toString());
builder.delete(0, builder.length());
builder.append(s.charAt(j));
} else {
builder.append(s.charAt(j));
}
}
//添加最后的builder内容
if (builder.length() != 0) {
cur.add(builder.toString());
}
backtracking(s);
}
return result;
}
public void backtracking(String s) {
for (String s1 : cur) {
if (!isPalindrome(s1)) {
cur.clear();
} else {
result.add(new ArrayList<String>(cur));
cur.clear();
}
return;
}
}
public boolean isPalindrome(String s) {
char[] chars = s.toCharArray();
for (int i = 0; i < chars.length / 2; i++) {
if (chars[i] != chars[chars.length - i - 1]) {
return false;
}
}
return true;
}
/**
* 自己的思路2：分别按不同长度对s进行划分，不是回溯法
* 速度击败77.01%，内存击败23.77%
*
* @param s
* @return
*/
public List<List<String>> partition1(String s) {
backtracking1(s);
return result;
}
public void backtracking1(String s) {
if (s.length() == 1) {
cur.add(s);
result.add(new ArrayList<String>(cur));
return;
}
for (int i = 1; i < s.length(); i++) {
//按长度分割
String str1 = s.substring(0, i);
cur.add(str1);
if (!isPalindrome(str1)) {
cur.remove(cur.size() - 1);
continue;
}
String str2 = s.substring(i, s.length());
backtracking1(str2);
cur.remove(cur.size() - 1);
cur.remove(cur.size() - 1);
}
cur.add(s);
if (isPalindrome(s)) {
result.add(new ArrayList<String>(cur));
return;
}
}
/**
* 代码随想录思路：分别按不同长度对s进行划分，不是回溯法，在代码上进行了优化，当前是回文再加入，避免了删除两次等逻辑
* 速度击败16.16%，内存击败26.93%
*
* @param s
* @return
*/
public List<List<String>> partition2(String s) {
backtracking2(s,0);
return result;
}
public void backtracking2(String s,int start) {
if (start>=s.length()) {
result.add(new ArrayList<String>(cur));
return;
}
for (int i = start; i < s.length(); i++) {
//按长度分割
if (isPalindrome1(s,start,i)) {
String str1 = s.substring(start, i+1);
cur.add(str1);
}else {
continue;
}
backtracking2(s,i+1);
cur.remove(cur.size() - 1);
}
}
public boolean isPalindrome1(String s, int start, int end) {
char[] chars = s.toCharArray();
for (int i = start, j = end; i < j; i++, j--) {
if (chars[i] != chars[j]) {
return false;
}
}
return true;
}
}

178
TensorFlow_eaxmple/Model_train_test/condition_monitoring/self_try/compare/RNet.py
View File

@ -22,6 +22,7 @@ from model.Joint_Monitoring.compare.RNet import Joint_Monitoring
from model.CommonFunction.CommonFunction import * from model.CommonFunction.CommonFunction import *
from sklearn.model_selection import train_test_split from sklearn.model_selection import train_test_split
from tensorflow.keras.models import load_model, save_model from tensorflow.keras.models import load_model, save_model
import random
'''超参数设置''' '''超参数设置'''
time_stamp = 120 time_stamp = 120
@ -35,7 +36,7 @@ K = 18
namuda = 0.01 namuda = 0.01
'''保存名称''' '''保存名称'''
save_name = "./model/weight/{0}_timestamp{1}_feature{2}_weight/weight".format(model_name, save_name = "./model/weight/{0}_timestamp{1}_feature{2}_weight_epoch2_loss0.007/weight".format(model_name,
time_stamp, time_stamp,
feature_num, feature_num,
batch_size, batch_size,
@ -234,53 +235,6 @@ def EWMA(data, K=K, namuda=namuda):
pass pass
def get_MSE(data, label, new_model):
predicted_data = new_model.predict(data)
temp = np.abs(predicted_data - label)
temp1 = (temp - np.broadcast_to(np.mean(temp, axis=0), shape=predicted_data.shape))
temp2 = np.broadcast_to(np.sqrt(np.var(temp, axis=0)), shape=predicted_data.shape)
temp3 = temp1 / temp2
mse = np.sum((temp1 / temp2) ** 2, axis=1)
print("z:", mse)
print(mse.shape)
# mse=np.mean((predicted_data-label)**2,axis=1)
print("mse", mse)
dims, = mse.shape
mean = np.mean(mse)
std = np.sqrt(np.var(mse))
max = mean + 3 * std
# min = mean-3*std
max = np.broadcast_to(max, shape=[dims, ])
# min = np.broadcast_to(min,shape=[dims,])
mean = np.broadcast_to(mean, shape=[dims, ])
# plt.plot(max)
# plt.plot(mse)
# plt.plot(mean)
# # plt.plot(min)
# plt.show()
#
#
return mse, mean, max
# pass
def condition_monitoring_model():
input = tf.keras.Input(shape=[time_stamp, feature_num])
conv1 = tf.keras.layers.Conv1D(filters=256, kernel_size=1)(input)
GRU1 = tf.keras.layers.GRU(128, return_sequences=False)(conv1)
d1 = tf.keras.layers.Dense(300)(GRU1)
output = tf.keras.layers.Dense(10)(d1)
model = tf.keras.Model(inputs=input, outputs=output)
return model
# trian_data:(300455,120,10) # trian_data:(300455,120,10)
# trian_label1:(300455,10) # trian_label1:(300455,10)
# trian_label2:(300455,) # trian_label2:(300455,)
@ -523,6 +477,94 @@ def showResult(step_two_model: Joint_Monitoring, test_data, isPlot: bool = False
return total_result return total_result
def get_MSE(data, label, new_model, isStandard: bool = True, isPlot: bool = True):
predicted_data1 = []
predicted_data2 = []
predicted_data3 = []
size, length, dims = data.shape
for epoch in range(0, size - batch_size + 1, batch_size):
each_test_data = data[epoch:epoch + batch_size, :, :]
output1, output2, output3, _ = new_model.call(inputs=each_test_data, is_first_time=True)
predicted_data1.append(output1)
predicted_data2.append(output2)
predicted_data3.append(output3)
predicted_data1 = np.reshape(predicted_data1, [-1, 10])
predicted_data2 = np.reshape(predicted_data2, [-1, 10])
predicted_data3 = np.reshape(predicted_data3, [-1, 10])
temp = np.abs(predicted_data1 - label)
temp1 = (temp - np.broadcast_to(np.mean(temp, axis=0), shape=predicted_data1.shape))
temp2 = np.broadcast_to(np.sqrt(np.var(temp, axis=0)), shape=predicted_data1.shape)
temp3 = temp1 / temp2
mse = np.sum((temp1 / temp2) ** 2, axis=1)
print("z:", mse)
print(mse.shape)
# mse=np.mean((predicted_data-label)**2,axis=1)
print("mse", mse)
if isStandard:
dims, = mse.shape
mean = np.mean(mse)
std = np.sqrt(np.var(mse))
max = mean + 3 * std
print("max:", max)
# min = mean-3*std
max = np.broadcast_to(max, shape=[dims, ])
# min = np.broadcast_to(min,shape=[dims,])
mean = np.broadcast_to(mean, shape=[dims, ])
if isPlot:
plt.figure(random.randint(1, 9))
plt.plot(max)
plt.plot(mse)
plt.plot(mean)
# plt.plot(min)
plt.show()
else:
if isPlot:
plt.figure(random.randint(1, 9))
plt.plot(mse)
# plt.plot(min)
plt.show()
return mse
return mse, mean, max
# pass
# healthy_data是健康数据,用于确定阈值，all_data是完整的数据,用于模型出结果
def getResult(model: tf.keras.Model, healthy_data, healthy_label, unhealthy_data, unhealthy_label, isPlot: bool = False,
isSave: bool = False):
# TODO 计算MSE确定阈值
mse, mean, max = get_MSE(healthy_data, healthy_label, model)
# 误报率的计算
total, = mse.shape
faultNum = 0
faultList = []
for i in range(total):
if (mse[i] > max[i]):
faultNum += 1
faultList.append(mse[i])
fault_rate = faultNum / total
print("误报率:", fault_rate)
# 漏报率计算
missNum = 0
missList = []
mse1 = get_MSE(unhealthy_data, unhealthy_label, model, isStandard=False)
all, = mse1.shape
for i in range(all):
if (mse1[i] < max[0]):
missNum += 1
missList.append(mse1[i])
miss_rate = missNum / all
print("漏报率:", miss_rate)
pass
if __name__ == '__main__': if __name__ == '__main__':
total_data = loadData.execute(N=feature_num, file_name=file_name) total_data = loadData.execute(N=feature_num, file_name=file_name)
total_data = normalization(data=total_data) total_data = normalization(data=total_data)
@ -535,38 +577,26 @@ if __name__ == '__main__':
# 单次测试 # 单次测试
# train_step_one(train_data=train_data_healthy[:256, :, :], train_label1=train_label1_healthy[:256, :], # train_step_one(train_data=train_data_healthy[:256, :, :], train_label1=train_label1_healthy[:256, :],
# train_label2=train_label2_healthy[:256, ]) # train_label2=train_label2_healthy[:256, ])
#### 模型训练
train_step_one(train_data=train_data_healthy, train_label1=train_label1_healthy, train_label2=train_label2_healthy) # train_step_one(train_data=train_data_healthy, train_label1=train_label1_healthy, train_label2=train_label2_healthy)
# 导入第一步已经训练好的模型,一个继续训练，一个只输出结果 # 导入第一步已经训练好的模型,一个继续训练，一个只输出结果
# step_one_model = Joint_Monitoring() step_one_model = Joint_Monitoring()
# step_one_model.load_weights(save_name) step_one_model.load_weights(save_name)
# #
# step_two_model = Joint_Monitoring() # step_two_model = Joint_Monitoring()
# step_two_model.load_weights(save_name) # step_two_model.load_weights(save_name)
# #### TODO 第二步训练 # #### TODO 计算MSE
# ### healthy_data.shape: (300333,120,10) healthy_size, _, _ = train_data_healthy.shape
# ### unhealthy_data.shape: (16594,10) unhealthy_size, _, _ = train_data_unhealthy.shape
# healthy_size, _, _ = train_data_healthy.shape all_data, _, _ = get_training_data_overlapping(
# unhealthy_size, _, _ = train_data_unhealthy.shape total_data[healthy_size - 2 * unhealthy_size:unhealthy_date, :], is_Healthy=True)
# # train_data, train_label1, train_label2, test_data, test_label1, test_label2 = split_test_data(
# # healthy_data=train_data_healthy[healthy_size - 2 * unhealthy_size:, :, :], getResult(step_one_model, healthy_data=train_data_healthy[healthy_size - 2 * unhealthy_size:, :],
# # healthy_label1=train_label1_healthy[healthy_size - 2 * unhealthy_size:, :], healthy_label=train_label1_healthy[healthy_size - 2 * unhealthy_size:, :],
# # healthy_label2=train_label2_healthy[healthy_size - 2 * unhealthy_size:, ], unhealthy_data=train_data_unhealthy, unhealthy_data=train_data_unhealthy, unhealthy_label=train_label1_unhealthy)
# # unhealthy_label1=train_label1_unhealthy, unhealthy_label2=train_label2_unhealthy)
# # train_step_two(step_one_model=step_one_model, step_two_model=step_two_model,
# # train_data=train_data,
# # train_label1=train_label1, train_label2=np.expand_dims(train_label2, axis=-1))
#
# ### TODO 测试测试集
# step_one_model = Joint_Monitoring()
# step_one_model.load_weights(save_name)
# step_two_model = Joint_Monitoring()
# step_two_model.load_weights(save_step_two_name)
# # test(step_one_model=step_one_model, step_two_model=step_two_model, test_data=test_data, test_label1=test_label1,
# # test_label2=np.expand_dims(test_label2, axis=-1))
#
# ###TODO 展示全部的结果 # ###TODO 展示全部的结果
# all_data, _, _ = get_training_data_overlapping( # all_data, _, _ = get_training_data_overlapping(
# total_data[healthy_size - 2 * unhealthy_size:unhealthy_date, :], is_Healthy=True) # total_data[healthy_size - 2 * unhealthy_size:unhealthy_date, :], is_Healthy=True)